Differentiability and continuity of a multivariable function 
Let $f:\mathbb R^2\to \mathbb R$ be defined by $$f(x,y)=\begin{cases}\frac{x|y|}{\sqrt{x^2+y^2}},& (x,y)\ne(0,0)\\ 0,& (x,y)=(0,0).\end{cases}$$
  
  
*
  
*For which non-zero vectors $u$ does the directional derivative exist at the point $(0,0)$?
  
*Do the partial derivatives exist?
  
*Is $f$ differentiable at $0$?
  
*Is $f$ continuous at $0$?
  

According to my calculations, the directional derivative is zero for all non-zero vectors $u$ at $(0,0)$ which implies the partial derivatives are zero at $(0,0)$. I am not sure how to proceed with 3. and 4. I am not allowed to use the existence of partial derivatives and their continuity to check for differentiability since the author introduces this theorem in the next section. 
Thanks.
 A: For 3. you should compute directly
$$
\lim_{\|(x,y) \| \to 0} \frac{f(x,y)}{\sqrt{x^2+y^2}}=\lim_{\substack{x \to 0 \\ y \to 0}} \frac{x|y|}{x^2+y^2}.
$$
Of course this can be done once you are sure that the partial derivatives at $(0,0)$ are both zero. By choosing the special path $x=y$, it is immediate to see that the limit is not zero, and therefore $f$ is not differentiable at $(0,0)$.
The continuity of $f$ at zero follows immediately from the inequality
$$
|y| \leq \sqrt{x^2+y^2},
$$
since
$$
|f(x,y)| \leq |x| \frac{|y|}{\sqrt{x^2+y^2}} \leq |x| \to 0
$$
as $x\to 0$, $y \to 0$.
A: $f$ is continuous at $(0,0)$ as you can prove the inequality $$ \vert f(x,y) \vert \le \frac{\vert xy \vert}{\sqrt{x^2+y^2}} \le \frac{\sqrt{x^2+y^2}}{2}$$
$f$ is not differentiable at $(0,0)$. If it would be the case its derivative would be equal to the zero matrix as the partial derivatives are equal to $0$. Hence all the directional derivatives would vanish which is not the case. A contradiction proving that $f$ is not differentiable at the origin.
